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Math Help - Integer numbers

  1. #1
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    Bogota D.D
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    Integer numbers

    Hi everyone.

    I can't show this exercise:

    Let a,b\in{\mathbb{Z} be and a<b then a+1\leq{b}

    I've tried by contraction, but I haven't gotten anything.

    Thanks
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  2. #2
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    Re: Integer numbers

    Well, I've done following:

    Suppose that a+1>b, but a<b\Leftrightarrow{a+1<b+1} hence it's a contraction because a+1>b and a+1<b+1.

    Therefore a+1\leq{b}

    Is it correct?
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  3. #3
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    Re: Integer numbers

    Quote Originally Posted by Fernando View Post
    Hi everyone.

    I can't show this exercise:

    Let a,b\in{\mathbb{Z} be and a<b then a+1\leq{b}

    I've tried by contraction, but I haven't gotten anything.

    Thanks
    Suppose :

    a+1 > b

    hence :

    a>b-1 \Rightarrow a \geq b

    so we have contradiction because :

    a < b
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  4. #4
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    Re: Integer numbers

    Quote Originally Posted by princeps View Post
    Suppose :

    a+1 > b

    hence :

    a>b-1 \Rightarrow a \geq b

    so we have contradiction because :

    a < b
    Thanks! But is my answer correcto ?
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  5. #5
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    Re: Integer numbers

    Quote Originally Posted by Fernando View Post
    Thanks! But is my answer correcto ?
    Since :

    a+1 > b ~\text{and}~ a+1 < b+1

    cannot be true at the same time your reasoning is correct .
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  6. #6
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    Re: Integer numbers

    OK! thanks again !
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  7. #7
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    Re: Integer numbers

    Quote Originally Posted by Fernando View Post
    Well, I've done following:
    Suppose that a+1>b, but a<b\Leftrightarrow{a+1<b+1} hence it's a contraction because a+1>b and a+1<b+1.
    Therefore a+1\leq{b}
    Is it correct?
    It is difficult to answer that. We don't even know what set of axioms are in use here.
    Here is a standard theorem in the development of number systems.
    Given any two numbers, if b-a>1 then (\exists j\in\mathbb{Z})[a<j<b].
    That depends upon Archimedes' principle and well ordering.
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